A community for students.

Here's the question you clicked on:

55 members online
  • 0 replying
  • 0 viewing

ganeshie8

  • one year ago

consider a regular \(n-gon\). show that a "flip", followed by a "ccw rotation" equals the "cw rotation" followed by a "flip": \[\large rf=fr^{-1}\]

  • This Question is Closed
  1. perl
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 4

    |dw:1440623798326:dw|

  2. perl
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 4

    That's a concrete example, to prove it abstractly ...

  3. ganeshie8
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 1

    |dw:1440623844373:dw|

  4. perl
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 4

    |dw:1440623936620:dw|

  5. perl
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 4

    so we want to show r * f = f * r^-1 , where * stands for composition we can use the fact that f*f = Identity

  6. ganeshie8
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 1

    i think so :) \(f = f^{-1}\) is kinda obvious as reflection undoes itself..

  7. anonymous
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 0

    https://en.wikipedia.org/wiki/Dihedral_group#Group_structure

  8. perl
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 4

    r * f = r *f^-1 = (r^-1)^-1 * f^-1 = ( f * r^-1 )^-1

  9. perl
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 4

    Apparently I proved that r*f and f*r^-1 are inverses, but that wasn't the question :)

  10. perl
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 4

    using (a*b)^-1 = b^-1 * a^-1

  11. ganeshie8
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 1

    Haha nice :) that is also clear from shoe socks principle \((rf)^{-1} = f^{-1}r^{-1} = fr^{-1}\)

  12. perl
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 4

    is there a proof r f = f r^-1 without having to go into the geometry , considering when n is even and odd, etc

  13. ganeshie8
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 1

    not so sure, i got stuck on this while attempting below exercises https://i.gyazo.com/6b1fbd9dc197477edffe9987ca355527.png

  14. ganeshie8
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 1

    they are from Gallian abstract algebra book

  15. perl
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 4

    You can use the fact that frf = r^-1

  16. perl
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 4

    http://prntscr.com/899tqo

  17. perl
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 4

    given f*f = I (f * r* f) = r^-1 r* f = I * (r*f) = (f* f )* r * f = f * ( f* r * f ) = f * r ^-1

  18. perl
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 4

    'given' meaning these are theorems for rigid transformations of n-gon

  19. perl
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 4

    f r f = r^-1 In geometric terms: in the mirror a rotation looks like an inverse rotation.

  20. ganeshie8
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 1

    that looks interesting!

  21. ganeshie8
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 1

    but aren't we trying to prove the same thing

  22. ganeshie8
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 1

    \[\large rf=fr^{-1}\] left multiplying both sides by \(f\) gives \[\large frf=r^{-1}\]

  23. perl
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 4

    |dw:1440626758905:dw|